Functions Worked Examples (Medium)

Q: How do I find a function hidden inside a composite?

Apply the known outer function to g(x)g(x)g(x) as if g(x)g(x)g(x) were its input, set that equal to the given composite, and solve for g(x)g(x)g(x). Always rebuild the composite at the end to check.

Q: Does f^{2}(x) mean I square f(x)?

No. In this chapter f2(x)=ff(x)=f(f(x))f^{2}(x)=ff(x)=f(f(x))f2(x)=ff(x)=f(f(x)), the function applied twice. So for f(x)=3x+5f(x)=3x+5f(x)=3x+5, f2(x)=9x+20f^{2}(x)=9x+20f2(x)=9x+20, not (3x+5)2(3x+5)^{2}(3x+5)2.

Q: How do I compare two expressions that must be equal?

If px+q=9x+20px+q=9x+20px+q=9x+20 holds for all xxx, then p=9p=9p=9 and q=20q=20q=20. Match the xxx-coefficients and the constant terms separately; each comparison gives you one equation.

What these examples cover

These medium Functions examples move beyond direct substitution into the reasoning the exam rewards most. You will work backwards from a composite to recover a hidden function, rearrange a rational expression to find an inverse, and use the self-composite $f^{2}(x)=ff(x)$ to pin down an unknown constant.

None of these needs heavy algebra, each needs a clear plan and a tidy rearrangement. Cover each solution, set the problem up yourself, and only compare once you have a final answer.

When you check, focus on the step that turns the question around, because that reversal is where most of the marks are won or lost. Every answer here is confirmed by an independent substitution.

Worked examples

Attempt all three fully before reading the solutions. Watch for the moment each question asks you to reverse a step you already know how to do forwards.

Q1[4 marks]

The function $f$ is defined by $f(x)=2x-1$ . Given that the composite function $fg(x)=6x+5$ , find $g(x)$ .

Show worked solution

$fg(x)$ means $f(g(x))$ , so $g$ is the inner function. Apply the rule of $f$ to the whole of $g(x)$ : wherever $f$ takes its input, put $g(x)$ .

fg(x)=2\,g(x)-1

Set this equal to the given composite and solve for $g(x)$ :

2\,g(x)-1=6x+5

2\,g(x)=6x+6

g(x)=3x+3

Answer

$g(x)=3x+3$ . Check by rebuilding the composite: $fg(x)=2(3x+3)-1=6x+6-1=6x+5$ , which matches the given expression exactly.

Q2[4 marks]

A function is defined by $f(x)=\dfrac{2x+1}{x-3}$ , $x\neq 3$ . Find $f^{-1}(x)$ and state the value of $x$ it excludes.

Show worked solution

Let $y=f(x)$ , then make $x$ the subject. Begin by clearing the denominator:

y=\frac{2x+1}{x-3} \;\Rightarrow\; y(x-3)=2x+1

Expand, gather every $x$ term on one side, and factorise out $x$ :

yx-3y=2x+1

yx-2x=3y+1

x(y-2)=3y+1

Divide to isolate $x$ , then rewrite in terms of $x$ to state the inverse:

x=\frac{3y+1}{y-2} \;\Rightarrow\; f^{-1}(x)=\frac{3x+1}{x-2}

Answer

$f^{-1}(x)=\dfrac{3x+1}{x-2}$ , which excludes $x=2$ (its denominator is zero there). Check with one value: $f(4)=\dfrac{9}{1}=9$ , and $f^{-1}(9)=\dfrac{28}{7}=4$ , returning the original input.

Q3[4 marks]

The function $f$ is defined by $f(x)=3x+k$ , where $k$ is a constant. Given that $f^{2}(x)=9x+20$ , find the value of $k$ .

Show worked solution

Here $f^{2}(x)$ means $ff(x)=f(f(x))$ , the function composed with itself, not $[f(x)]^{2}$ . Substitute $f(x)=3x+k$ into $f$ once more:

f^{2}(x)=f(3x+k)=3(3x+k)+k

Expand the bracket and collect like terms:

f^{2}(x)=9x+3k+k=9x+4k

Match this with the given $9x+20$ . The $x$ -terms already agree, so compare the constant terms:

4k=20 \;\Rightarrow\; k=5

Answer

$k=5$ . Check: with $f(x)=3x+5$ , $f^{2}(x)=3(3x+5)+5=9x+15+5=9x+20$ , as required.

Q4[4 marks]

The functions $f$ and $g$ are defined by $f(x)=x+1$ and $g(x)=2x^{2}-3$ . Find $gf(x)$ in its simplest form, then find the values of $x$ that satisfy $gf(x)=15$ .

Show worked solution

$gf(x)$ means $g(f(x))$ , so substitute the whole of $f(x)$ into $g$ wherever $x$ appears, then expand:

gf(x)=2(x+1)^{2}-3=2x^{2}+4x-1

Set this equal to 15 and rearrange into $ax^{2}+bx+c=0$ , then factorise:

2x^{2}+4x-1=15 \;\Rightarrow\; x^{2}+2x-8=0 \;\Rightarrow\; (x+4)(x-2)=0

x+4=0 \;\Rightarrow\; x=-4 \qquad \text{or} \qquad x-2=0 \;\Rightarrow\; x=2

Answer

$x=-4$ or $x=2$ . Check: $f(-4)=-3$ and $g(-3)=2(9)-3=15$ ; $f(2)=3$ and $g(3)=2(9)-3=15$ , so both roots satisfy the equation.

Q5[3 marks]

The function $f$ is defined by $f(x)=\dfrac{3x-1}{2}$ . Find the value of $x$ for which $f^{-1}(x)=5$ .

Show worked solution

You do not need the general formula for $f^{-1}$ here. Since $f^{-1}(x)=5$ means $x$ is the image of 5 under $f$ , apply $f$ directly to 5:

x=f(5)=\dfrac{3(5)-1}{2}

x=\dfrac{14}{2}=7

Answer

$x=7$ . Check with the general inverse: rearranging $y=\dfrac{3x-1}{2}$ gives $f^{-1}(x)=\dfrac{2x+1}{3}$ , and $f^{-1}(7)=\dfrac{15}{3}=5$ , confirming the shortcut.

Q6[4 marks]

The function $f$ is defined by $f(x)=x^{2}-4x+6$ for the domain $-1\le x\le 4$ . Find the range of $f$ .

Show worked solution

Complete the square to locate the turning point, since that gives the smallest value the function can take:

f(x)=(x-2)^{2}+2

The minimum value is 2, occurring at $x=2$ , which lies inside the given domain. Because the graph opens upward, the largest value comes from whichever endpoint is farther from $x=2$ ; check both:

f(-1)=1+4+6=11 \qquad f(4)=16-16+6=6

Answer

Since $x=-1$ is 3 units from the turning point and $x=4$ is only 2 units away, the maximum is $f(-1)=11$ . The range is $2\le f(x)\le 11$ .

Q7[4 marks]

The function $f$ is defined by $f(x)=ax+b$ , where $a$ and $b$ are constants. Given that $f(1)=4$ and $f(4)=13$ , find the value of $a$ and of $b$ .

Show worked solution

Substitute each given input into $f(x)=ax+b$ to form a pair of simultaneous equations:

a+b=4 \qquad 4a+b=13

Subtract the first equation from the second to eliminate $b$ , then back-substitute:

3a=9 \;\Rightarrow\; a=3 \;\Rightarrow\; b=4-3=1

Answer

$a=3$ and $b=1$ . Check: $f(4)=3(4)+1=13$ , matching the given value.

Q8[3 marks]

The functions $f$ and $g$ are defined by $f(x)=2x-5$ and $g(x)=\dfrac{x+p}{2}$ , where $p$ is a constant. Given that $fg(x)=x$ for all values of $x$ , find the value of $p$ .

Show worked solution

$fg(x)=x$ for every $x$ is exactly the condition that $g$ undoes $f$ , so $g$ must be $f^{-1}$ . Build $fg(x)$ by substituting $g(x)$ into $f$ :

fg(x)=2\left(\dfrac{x+p}{2}\right)-5=x+p-5

For this to equal $x$ for every value of $x$ , the constant term must vanish:

p-5=0 \;\Rightarrow\; p=5

Answer

$p=5$ , so $g(x)=\dfrac{x+5}{2}$ is exactly $f^{-1}(x)$ . Check with one value: $f(3)=1$ and $g(1)=\dfrac{1+5}{2}=3$ , returning the original input.

The thread running through all three is reversal. In the first you undo a composite to expose an inner function; in the second you undo a function to build its inverse; in the third you read a self-composite backwards to a single unknown.

Once you recognise that a question is really asking you to run a familiar step in reverse, the plan almost writes itself.

Key method points

These three examples share one habit: turn the question around and rearrange cleanly. Keep these points to hand as you practise.

To recover a hidden function from $fg(x)$ , apply the outer rule to $g(x)$ , set the result equal to the given composite, then solve.
For an inverse, write $y=f(x)$ , make $x$ the subject, then swap to $x$ ; the value the inverse excludes is where its denominator is zero.
Read $f^{2}(x)$ as $ff(x)$ , the function applied twice, never as $[f(x)]^{2}$ .
When two expressions in $x$ must be equal for all $x$ , compare the $x$ -coefficients and the constants separately.
Confirm every answer with one clean substitution, it is the quickest guard against a slipped sign or a dropped bracket.

How a teacher helps

Medium Functions questions reward a clear plan more than clever algebra, and that is exactly what a lesson builds. Our teacher watches how you set the problem up, which function is inner, how you make $x$ the subject, and steps in the moment the plan goes astray, so you are not left rearranging in circles.

Because our teachers are experienced, you get someone who can show a second route when the first one feels stuck. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation stays familiar whichever version you sit.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

How do I find a function hidden inside a composite?

Apply the known outer function to $g(x)$ as if $g(x)$ were its input, set that equal to the given composite, and solve for $g(x)$ . Always rebuild the composite at the end to check.

What value does an inverse function exclude?

The inverse is undefined where its own denominator is zero. For $f^{-1}(x)=\dfrac{3x+1}{x-2}$ , that is $x=2$ .

This excluded value corresponds to a boundary of the original function's range.

Does

f^{2}(x)

mean I square

f(x)

No. In this chapter $f^{2}(x)=ff(x)=f(f(x))$ , the function applied twice.

So for $f(x)=3x+5$ , $f^{2}(x)=9x+20$ , not $(3x+5)^{2}$ .

How do I compare two expressions that must be equal?

If $px+q=9x+20$ holds for all $x$ , then $p=9$ and $q=20$ . Match the $x$ -coefficients and the constant terms separately; each comparison gives you one equation.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Functions, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

Ready to get started?